Refer to the diagram shown below.
The piston supports the same load W at both temperatures.
The ideal gas law is
[tex]pV=nRT[/tex]
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant
State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter
State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter
Because V, n, and R remain the same between the two temperatures, therefore
[tex] \frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}} [/tex]
If the supported load is W kg, then
[tex]p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa[/tex]
Similarly,
[tex]p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa[/tex]
[tex] \frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}[/tex]
Because p₁/p₂ = T₁/T₂, therefore
[tex]16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m[/tex]
The minimum piston diameter at 150 C is 20.8 cm.
Answer: 20.8 cm diameter
